Question

Compute the nth derivative f^{(n)}(t) of the function f(t)=frac{(e^t-1)}{t}. Then compute \[\lim_{t \rightarrow 0}f^{(n)}(t)\]. This is from probability theory dealing with moment generating functions. I'm having trouble solving this.

Answer 1

yuck the latex came out badly

Answer 2

let me re-write

Answer 3

find the nth derivative of:
\[\frac{e^t-1}{t}\]

Answer 4

with respect to t, and then take the limit of that nth derivative as t goes to 0

Answer 5

If I was going to attempt it. I would take like 4 derivative as it stands and see if you can develop a pattern? From there, the limit will fall out.

Answer 6

i took a bunch of derivatives, but there are new terms each time so I think you have to write the nth derivative as a series

Answer 7

Let me try something. h/o

Answer 8

ok

Answer 9

hey male sory to interupt but can u check my question please...@ least where i got it sof ar

Answer 10

Yeah, give me like 2 min please

Answer 11

ok no prob

Answer 12

purple did you post your question maybe I can help

Answer 13

Well, vitale, I got half of it in terms of an nth derivative:
\[\frac{d^n}{dt^n}(\frac{e^t-1}{t})=\frac{e^t(t-n)}{t^{2^n}}+(-1)^{n+1}...\]. However, I can't figure out how to handle the coefficient or powers of t on the second term. Take the first 3 derivatives. And see if what I have is right for the first part plesae.

Answer 14

Look at that, I'll help purp.

Answer 15

ok

Answer 16

i had maple compute the first 9, and I dont see your first term in the derivatives

Answer 17

I might have it.
Try this:
\[\frac{d^n}{dt^n}(\frac{e^t-1}{t})=\frac{e^t(t-n)}{t^{2^n}}+(-1)^{n+1}(n!)t^{1(n+1)} \].
Really? At least, on through 3 these work. I broke it up and differentiated e^t/t -t^-1. Then made a pattern. Try it by hand split up. Plug in your n's and see if you get it? :/ This is the best I can do.

Answer 18

If you refresh the page that won't look like pellet >.>

Answer 19

let me see if it simplifies to that, one sec

Answer 20

And in the last term it should be t^-(n+1) not a 1

Answer 21

That works PERFECTLY for me :D

Answer 22

it's not working for first derivative for me

Answer 23

oh woops yes it is

Answer 24

let me check the higher ones

Answer 25

If you plug in 1 to my equation you get:
\[\frac{e^t(t-1)}{t^2}+t^{-2}\].
Okay, I did up through 4 and they all worked. But double check it :P

Answer 26

ok let me look at it for a bit

Answer 27

maple is saying only the first derivative works

Answer 28

the other ones are missing terms

Answer 29

what did you compute as the 2nd or 3rd derivative?

Answer 30

Really? Hmmm...Well for the second derivative my equation gives:
\[\frac{e^t(t-2)}{t^4}-2t^{-3}\]. Which is what I get doing it by hand. It should only be 2 terms if you don't break up the fraction.
Well the first derivative is:
\[\frac{e^t(t-1)}{t^2}+t^{-2}\].
The derivative of that (the 2nd of the original is:
\[\frac{e^t(t-1)-e^t(1)}{t^4}-2t^{-3}=\frac{e^t(t-2)}{t^4}-2t^{-3}\].
That fits doesn't it?

Answer 31

let me do it by hand and check

Answer 32

Okay.

Answer 33

this is looks like it is working out nasty like
lol
brb

Answer 34

for the second derivative I get:
\[\frac{e^t(t^2-2t+2)-2}{t^3}\]

Answer 35

LOL I KNOW WHY. I"M RETARDED. H/o let me retry xPPPPPP

Answer 36

Sorry for the wasted time D:

Answer 37

no problem im still working on it too.

Answer 38

For the second I get:
\[\frac{e^t(t^3-t^2-2t+1)}{t^4}\].

Answer 39

this is maple's nth deriv:

\[-\frac{-\left( \frac{1}{t}\right)^n (\Gamma(1+n)-\Gamma(1+n,-t))}{t}\]'

Answer 40

k let me check my second deriv

Answer 41

i got the same as before and just checked it on maple.

Answer 42

Damn, thats an intense nth derivative. lol. GAMMA FUNCTION!!!

Answer 43

yeah haha I don't know what to do with it as far as taking a limit

Answer 44

I see what you mean. Limits of series's are tricky...

Answer 45

well the solution to the limit should be 1/(n+1) since it's a known result in probability